设x,y,z>0,且x^2+y^2+z^2=1,求证x^2/(1+9xy)+y^2/(1+9xz)+z^2/(1+9xy)≥1/4

Question

设x,y,z>0,且x^2+y^2+z^2=1,求证 x^2/(1+9xy)+y^2/(1+9xz)+z^2/(1+9xy)≥1/4

Answer 1

题目有误吧,应该是x^2/(1+9xy)+y^2/(1+9yz)+z^2/(1+9zx)>=1/4
2010年衢州二中高三第二次高考模拟测试自选模块的不等式!
出题人是衢州二中舒金根老师,为一牛人.这是我们的二模卷.我给出我的
证明：由柯西不等式：
[(1+9xy)+(1+9yz)+(1+9zx)][x^2/(1+9xy)+y^2/(1+9yz)+z^2/(1+9zx)]>=(x+y+z)^2
上式也即x^2/(1+9xy)+y^2/(1+9yz)+z^2/(1+9zx)]>=(x+y+z)^2/[3+9(xy+yz+zx)]
注意到：因为x^2+y^2+z^2=1,所以(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=1+2(xy+yz+zx)
于是x^2/(1+9xy)+y^2/(1+9yz)+z^2/(1+9zx)]>=[1+2(xy+yz+zx)]/[3+9(xy+yz+zx)]
只要证明[1+2(xy+yz+zx)]/[3+9(xy+yz+zx)]>=1/4即可
而上式等价于：xy+yz+zx